چون $\frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{1}{3}$ و $\hat A = {90^ \circ }$ مشترک است پس دو مثلث $ABC$ و $ADE$ متشابه‌اند در نتیجه:

$\left\{ {\begin{array}{*{20}{c}}
{\hat D = \hat B}\\
{\hat E = \hat C}
\end{array}} \right.$

$\cos \hat E = \frac{3}{5}{\cos ^2}\theta  + {\sin ^2}\theta  = 1 \to \sin \hat E = \frac{4}{5}$
$ \Rightarrow \tan \hat E = \frac{{AD}}{{AE}} = \frac{{\sin \hat E}}{{\cos \hat E}} = \frac{{\frac{4}{5}}}{{\frac{3}{5}}} = \frac{4}{3}$
$ \Rightarrow \frac{{AD}}{{AE}} = \frac{4}{3} \Rightarrow \frac{{AE}}{{AD}} = \frac{3}{4}$
$\left\{ {\begin{array}{*{20}{c}}
{\tan \hat B = \tan \hat D = \frac{{AE}}{{AD}} = \frac{3}{4}}\\
{\sin \hat C = \sin \hat E = \frac{4}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}} \right. \Rightarrow \frac{3}{4} + \frac{4}{5} = \frac{{15 + 16}}{{20}} = \frac{{31}}{{20}}$